Time period of a pendulum

For a simple pendulum, $T = 2\pi \sqrt{\frac{l}{g}}$

Hence, $T \propto \sqrt{l}$

Relate fractional change in period to fractional change in length

Taking fractional changes, $\frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta l}{l}$

Given the length increases by $0.1\%$, $\frac{\Delta l}{l} = 0.1\% = \frac{0.1}{100} = 0.001$

Therefore, $\frac{\Delta T}{T} = \frac{1}{2}(0.001) = 0.0005$

So the time period increases by $0.05\%$.

Effect on clock

If the pendulum length increases, the time period increases, so the clock runs slow.

Fractional error in time per day: $0.0005 \times 24 \times 60 \times 60$

Since one day has $86400\,\text{s}$, $\text{error per day} = 0.0005 \times 86400 = 43.2\,\text{s}$

Final answer

The clock loses $\boxed{43.2\,\text{s per day}}$

Option check

• A: $86.4\,\text{s}$ ❌
• B: $4.32\,\text{s}$ ❌
• C: $43.2\,\text{s}$ ✅
• D: $8.64\,\text{s}$ ❌